Maths Skills 2 (OCR A-Level Biology A): Revision Notes

Maths Skills 2

Introduction

This chapter covers advanced mathematical skills required for A-Level Biology, focusing on two main areas:

• Using logarithms and log graphs to display and analyse data
• Applying statistical methods to analyse experimental results

These skills build upon the mathematical foundation established in Chapter 18 of OCR A Level Biology 1 Student's Book.

Variables in biological investigations

In all biological investigations, different types of variables are identified, measured or controlled. Understanding these categories is essential for experimental design and data analysis.

Independent variable – This is the variable you deliberately manipulate during your investigation. You select its range and the specific values to test. For example, when investigating photosynthesis, you might choose different light intensities or carbon dioxide concentrations as your independent variable. The independent variable is always plotted on the x-axis of a graph. Some investigations may have more than one independent variable.

Dependent variable – This is the variable you count or measure in response to changes in the independent variable. It represents the outcome or result of your investigation. For instance, you might measure the rate of photosynthesis or the biomass of organisms. Multiple dependent variables can be recorded in a single investigation.

Derived variable – This is a variable calculated from your collected results rather than directly measured. For example, rates of respiration or photosynthesis are often derived variables. You might collect data on bubble movement or gas volume over time, then calculate the rate from these measurements. Derived variables frequently appear as the final output in investigations.

Controlled variable – Any variable kept constant throughout an investigation to ensure a fair test. For example, temperature might be controlled when investigating the effect of light intensity on photosynthesis. Controlled variables prevent confounding factors from affecting your results.

Logarithms and log graphs

When to use logarithmic scales

Logarithmic scales become necessary when your independent or dependent variable covers an extremely wide range, making it difficult to plot data on standard linear graph paper. A logarithmic scale uses intervals that increase by an order of magnitude ($\times 10$), allowing you to display data spanning several powers of ten.

Common situations requiring logarithmic scales include:

• Plant hormone concentrations ranging from $0.0001\%$ to $1.0\%$
• Microorganism population growth, which increases exponentially
• Serial dilution experiments

When the values are exact orders of magnitude (e.g., $0.0001$, $0.001$, $0.01$, $0.1$, $1$), these are plotted equidistant on the axis. Intermediate values are accommodated between these major divisions.

Understanding log graph paper

Log graph paper consists of cycles, where each cycle represents one order of magnitude. For example:

• $0.1$ to $1.0$ = one cycle
• $1.0$ to $10.0$ = one cycle
• $10.0$ to $100.0$ = one cycle

Plotting data on log graphs

You have two options for creating log graphs:

1. Direct plotting – Plot your raw data directly onto log graph paper
2. Calculate logarithms first – Convert your data to logarithms (base 10), then plot on standard linear graph paper

The second method is demonstrated in the following example using bacterial growth data.

Converting between logs and actual numbers

To convert from a logarithm back to the actual number:

• Use the $10^x$ button on your calculator, where $x$ is the logarithm value
• In Excel, enter =Power(10,A1) where A1 contains the logarithm

This conversion is essential when extrapolating from log graphs to predict future values, such as pest populations in crop protection.

Example: Straightening the curve

When bacterial populations grow under ideal conditions (adequate nutrients, optimal temperature, no limiting factors), they exhibit exponential growth, with the population doubling at regular intervals.

The table shows bacterial growth over ten hours, with the population doubling every 30 minutes. When plotted on standard linear graph paper, exponential growth produces a characteristic J-shaped curve that makes early changes difficult to see and extrapolation challenging.

Types of data

Understanding data types is crucial for selecting appropriate display methods and statistical tests. Data can be classified as either qualitative or quantitative.

Qualitative data

Qualitative data describe characteristics that cannot be measured numerically. They fall into two categories:

Categoric (nominal) data – These place observations into distinct categories with no inherent order. Examples include:

• Different phenotypes in genetic studies (e.g., tall vs dwarf plants)
• Presence or absence of roots on cuttings
• Different species or taxa

Display methods: Bar charts, pie charts

Ordered (ordinal) data – These have a ranking or order but lack precise numerical intervals. Examples include:

• Abundance scales in fieldwork (ACFOR, DAFOR scales)
• Arbitrary scales from $0$ to $10$
• Ratings or grades

Display methods: Bar charts

Quantitative data

Quantitative data consist of numerical measurements and can be further divided:

Continuous data – These can take any value within a range and are measured using instruments. Examples include:

• Rates of respiration, photosynthesis, or enzyme-catalysed reactions (e.g., $1.0$, $2.5$, $3.7$ $\mu\text{mol} \cdot \text{s}^{-1}$)
• Length, height, mass, temperature
• Concentrations

Display methods: Line graphs, histograms, or bar charts depending on the independent variable

Discrete data – These consist of whole numbers (integers) only and are obtained by counting. Examples include:

• Number of heartbeats per minute
• Number of plants in a quadrat
• Number of offspring with particular phenotypes

Display methods: Bar charts, histograms, or line graphs depending on the investigation

The table below summarises these data types:

Descriptive statistics

Descriptive statistics summarise and simplify data to make it more interpretable. They include measures of central tendency (mean, mode, median) and measures of dispersion (range, standard deviation).

Measures of central tendency

• Mean ($\bar{x}$) – The arithmetic average, calculated by summing all values and dividing by the number of observations
• Mode – The most frequently occurring value
• Median – The middle value when data are arranged in order

For biological data, the mean is most commonly used and reported.

Standard deviation

Standard deviation (SD) quantifies the spread or dispersion of data around the mean. A small standard deviation indicates that data points cluster tightly around the mean, while a large standard deviation shows greater variability.

Key properties of standard deviation:

• Approximately 64% of values lie within $\pm 1$ SD of the mean
• Approximately 95% of values lie within $\pm 2$ SD of the mean
• Approximately 99.7% of values lie within $\pm 3$ SD of the mean

Standard error of the mean

Standard error of the mean (SEM) provides information about the precision of the sample mean as an estimate of the population mean. Unlike standard deviation, which describes the variability within your sample, SEM indicates how confident you can be that your sample mean reflects the true population mean.

The formula for calculating SEM is:

$\text{SEM} = \frac{\text{SD}}{\sqrt{n}}$

where SD is the standard deviation and $n$ is the number of observations.

A smaller SEM indicates greater confidence in the relationship between the sample mean and population mean.

95% confidence intervals

The 95% confidence limits (95% CL) define a range within which we can be 95% certain that the true population mean lies. They are calculated using:

$95\% \text{ CL} = \bar{x} \pm 2 \times \text{SE}$

The confidence interval is the difference between the upper and lower confidence limits. These confidence limits provide the best measure of uncertainty for error bars on graphs and are superior to standard deviation for comparing different data sets.

Worked example: Photosynthesis rates

An investigation measured the rate of photosynthesis in Canadian pondweed (Elodea canadensis) under blue and green light. The investigation was replicated ten times.

Interpreting error bars

When comparing two sets of data using 95% confidence limit error bars:

• If error bars do not overlap – The difference between means is likely statistically significant
• If error bars overlap – The difference between means is definitely not statistically significant

Statistical tests

Biological data are inherently variable due to genetic differences, environmental factors, and measurement uncertainty. Statistical tests allow us to determine whether observed differences or relationships are likely due to real biological effects or simply chance variation.

Types of investigations requiring statistical tests

Biological investigations at A-Level typically fall into four categories:

1. Comparing observed vs expected results (e.g., genetic ratios) – Use chi-squared test
2. Testing for correlation between two variables (e.g., light intensity and plant density) – Use Spearman's rank correlation test
3. Comparing means of two samples (e.g., plant growth under different conditions) – Use Student's t-test
4. Comparing distributions (e.g., leaf lengths in different habitats) – Use Student's t-test

The chi-squared ($\chi^2$) test

The chi-squared test is a "goodness-of-fit" test that determines whether observed results differ significantly from expected results. It is exclusively used with categoric data.

When to use the chi-squared test:

• Data are categoric (nominal or ordinal)
• You can predict expected outcomes from theory (e.g., Mendelian ratios) or by assuming random distribution
• All expected values are 5 or greater
• Working with actual counts, not percentages

Formula:

$\chi^2 = \sum \frac{(O - E)^2}{E}$

where $\Sigma$ means sum of, $O$ is the observed value, and $E$ is the expected value.

Steps for performing a chi-squared test:

1. Identify the inheritance pattern – Determine if it's monohybrid, dihybrid, sex-linked, etc.
2. Determine the expected ratio – Use genetic theory to predict the expected ratio (e.g., 3:1, 9:3:3:1, 1:1:1:1)
3. Write a null hypothesis – For example: "There is no significant difference between the observed and expected results"
4. Create a chi-squared table with six columns:
   • Category
   • Observed (O)
   • Expected (E)
   • $O - E$
   • $(O - E)^2$
   • $\frac{(O - E)^2}{E}$
5. Complete the table:
   • Calculate expected values using the predicted ratio
   • Calculate the difference ($O - E$) for each category
   • Square each difference to remove negative signs
   • Divide each squared difference by the expected value
   • Sum the final column to obtain $\chi^2$
6. Calculate degrees of freedom: df = number of categories $- 1$
7. Find the critical value at $p = 0.05$ for your degrees of freedom using Table 15.4
8. Compare your calculated $\chi^2$ with the critical value:
   • If $\chi^2 \geq \chi^2_{\text{crit}}$: Reject null hypothesis (significant difference)
   • If $\chi^2 < \chi^2_{\text{crit}}$: Accept null hypothesis (no significant difference)
9. Write a conclusion stating:
   • The calculated $\chi^2$ value
   • The critical value
   • Whether you accept or reject the null hypothesis
   • What this means for the biological principle being tested

Interpreting probability values:

• $p > 0.90$ – Result looks suspicious (too good to be true)
• $p \geq 0.05$ – No significant difference from expected
• $p < 0.05$ – Significantly different from expected
• $p < 0.01$ – Highly significant difference
• $p < 0.001$ – Very highly significant difference

Example: Fruit fly wing length

Pure-breeding fruit flies with long wings were crossed with pure-breeding flies with vestigial (very small) wings. All F₁ offspring had long wings. The F₁ flies were then interbred.

F₂ results:

• Long wings: 390
• Vestigial wings: 114
• Total: 504

Expected ratio: 3:1 (from monohybrid inheritance)

Null hypothesis: There is no significant difference between observed and expected numbers.

Expected values:

• Long wings: $\frac{3}{4} \times 504 = 378$
• Vestigial wings: $\frac{1}{4} \times 504 = 126$

Spearman's rank correlation test

The Spearman's rank correlation coefficient ($r_s$) measures the strength and direction of association between two variables. It can be used with both ordinal and interval data, and the relationship does not need to be linear.

When to use Spearman's rank correlation test:

• Data points within samples are independent of each other
• Data are paired (e.g., soil moisture and plant abundance at the same sampling stations)
• You have ordinal or interval data
• A scattergraph suggests a possible increasing or decreasing relationship (but not one that increases then decreases)
• Between 5 and 30 paired observations (though more is acceptable)
• Individuals selected randomly from a population

Formula:

$r_s = 1 - \frac{6 \times \sum D^2}{n^3 - n}$

where $\sum D^2$ is the sum of squared differences between ranks and $n$ is the number of pairs of observations.

Steps for performing Spearman's rank correlation test:

1. Plot a scattergraph to visually assess whether a correlation may exist. The relationship could be positive, negative, or absent.
2. Write a null hypothesis – Usually: "There is no significant correlation between [variable 1] and [variable 2]"
3. Rank each data set – Assign rank 1 to the largest value in each set. If you prefer, rank from smallest to largest, but remain consistent across both data sets. For tied values, assign the average of the ranks they would have occupied.
4. Calculate the difference ($D$) between ranks for each pair by subtraction
5. Square each difference to give $D^2$, removing negative signs
6. Sum all $D^2$ values to calculate $\sum D^2$
7. Calculate $r_s$ by substituting $\sum D^2$ and $n$ into the formula
8. Interpret the value of $r_s$:
   • $r_s = +1$ indicates perfect positive correlation
   • $r_s = -1$ indicates perfect negative correlation
   • $r_s = 0$ indicates no correlation
   • Intermediate values show the strength of correlation
   • The sign (+ or −) indicates direction; ignore the sign when comparing to critical values
9. Find the critical value at $p = 0.05$ for your number of paired measurements:

1. Compare $|r_s|$ with the critical value (ignore the sign):
   • If $|r_s| \geq$ critical value: Reject null hypothesis (significant correlation)
   • If $|r_s| <$ critical value: Accept null hypothesis (no significant correlation)
2. Write a conclusion including:
   • Type of correlation (positive or negative)
   • Strength of correlation (the $r_s$ value)
   • Statistical significance level
   • Remember: correlation does not prove causation

Example: Dissolved solids and macroinvertebrate diversity

Researchers investigated whether total dissolved solids (TDS) correlate with macroinvertebrate diversity in the River Wye by sampling 13 stations.

Null hypothesis: There is no significant correlation between total dissolved solids and the number of macroinvertebrate taxa.

The scattergraph shows a clear positive relationship, suggesting correlation is likely.

Ranking and calculations:

Student's t-test for unpaired data

The t-test for unpaired data determines whether the means of two independent samples differ significantly. It assumes both samples show normal (bell-shaped) distribution.

When to use Student's t-test for unpaired data:

• Interval data collected from two separate populations
• Both data sets show approximately normal distribution
• Sample sizes are relatively small (typically < 30 in each sample)
• Sample sizes may differ between the two groups
• You want to determine if a difference exists between population means

Formula:

$t = \frac{|\bar{x} - \bar{y}|}{\sqrt{\frac{s_x^2}{n_x} + \frac{s_y^2}{n_y}}}$

where $\bar{x}$ and $\bar{y}$ are the sample means, $s_x$ and $s_y$ are the standard deviations, and $n_x$ and $n_y$ are the sample sizes.

Degrees of freedom:

$\text{df} = (n_x - 1) + (n_y - 1)$

Steps for performing Student's t-test for unpaired data:

1. Write a null hypothesis – Usually: "There is no significant difference between the mean of [sample X] and the mean of [sample Y]"
2. Calculate the mean of each data set
3. Calculate the standard deviation for each data set
4. Square the standard deviations to get $s^2$
5. Apply the formula:
   • Find the absolute difference between means (ignore the sign)
   • Square each SD and divide by the corresponding sample size
   • Add these two results and take the square root
   • Divide the difference in means by this result to obtain $t$
6. Calculate degrees of freedom using the formula above
7. Find the critical value at $p = 0.05$ for your degrees of freedom using Table 15.10
8. Compare calculated $t$ with critical value:
   • If $t \geq t_{\text{crit}}$: Reject null hypothesis (significant difference)
   • If $t < t_{\text{crit}}$: Accept null hypothesis (no significant difference)
9. Write a conclusion stating:
   • The calculated $t$ value
   • Whether you accept or reject the null hypothesis
   • The probability level
   • What this means biologically

Example: Limpet shell shape variation

Students investigated variation in the common limpet (Patella vulgata) on exposed and sheltered shores. They randomly sampled 10 limpets from each location and calculated length:height ratios.

Null hypothesis: There is no significant difference between length:height ratios on sheltered and exposed shores.

Student's t-test for paired data

The t-test for paired data is used when two measurements are taken from the same individuals, such as before-and-after studies. This test accounts for individual variation between subjects.

When to use Student's t-test for paired data:

• Two sets of interval data collected from the same individuals
• Before-and-after comparisons (e.g., effect of exercise on heart rate)
• Matched pairs or repeated measures designs

Formula:

$t = \frac{\bar{d}\sqrt{n}}{s_d}$

where $\bar{d}$ is the mean of the differences, $s_d$ is the standard deviation of differences, and $n$ is the number of individuals.

Degrees of freedom: $n - 1$

Steps for performing Student's t-test for paired data:

1. Write a null hypothesis – For example: "There is no significant difference between [measurement before] and [measurement after]"
2. Calculate the difference between paired measurements for each individual
3. Calculate the mean difference ($\bar{d}$)
4. Calculate the standard deviation of the differences ($s_d$)
5. Apply the formula to calculate $t$
6. Calculate degrees of freedom ($n - 1$)
7. Find the critical value at $p = 0.05$ for your degrees of freedom
8. Compare calculated $t$ with critical value and draw conclusions as before

Example: Effect of exercise on heart rate

A student investigated how skipping affects heart rate. Ten volunteers of similar age, sex, BMI and fitness wore heart rate monitors, and their pulse was recorded before and after 30 seconds of skipping.

Null hypothesis: There is no significant difference between heart rates before and after exercise.

Choosing the appropriate statistical test

Selecting the correct statistical test depends on your data type and research question. Use this decision framework:

Decision criteria

1. For categoric data:

• Use chi-squared test only
• Compares observed data with predicted or expected values
• Common in genetics experiments testing Mendelian ratios

2. For comparing means of two samples:

• Use Student's t-test if both datasets show normal distribution
• Choose unpaired t-test when data come from two different populations
• Choose paired t-test when two measurements come from the same individuals

3. For investigating correlation:

• Use Spearman's rank correlation test
• Works with ordinal or interval data
• Determines if two variables are associated

Summary table

Research Question	Data Type	Statistical Test
Do observed results match predicted ratios?	Categoric	Chi-squared test
Are two population means different?	Interval (normal distribution)	t-test for unpaired data
Do before/after measurements differ?	Interval (same individuals)	t-test for paired data
Are two variables correlated?	Ordinal or interval	Spearman's rank correlation

Remember!